Question: The equation of an ellipse $E$ is $\dfrac{y^2}{49}+\dfrac {(x-6)^{2}}{81} = 1$. What are its center $(h, k)$ and its major and minor radius?
Solution: The equation of an ellipse with center $(h, k)$ is $ \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1$ We can rewrite the given equation as $\dfrac{(x - 6)^2}{81} + \dfrac{(y - 0)^2}{49} = 1 $ Thus, the center $(h, k) = (6, 0)$ $81$ is bigger than $49$ so the major radius is $\sqrt{81} = 9$ and the minor radius is $\sqrt{49} = 7$.